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if the algebra of functions on a manifold is a Poisson ring then the manifold is symplectic
Let $M$ be a smooth manifold and let $A$ be the algebra of smooth functions from $M$ to $\mathbb{R}$. Suppose that there exists a bilinear operation $[,]\colon A\times A\to A$ which makes $A$ a Poisson ring.
For this proof, we shall use the fact that $T^{*}(M)$ is the sheafification of the $A$module generated by the set $\{dff\in A\}$ modulo the relations

$d(f+g)=df+dg$

$dfg=g\,df+f\,dg$
Let us define a map $\omega\colon T^{*}(M)\to T(M)$ by the following conditions:

$\omega(df)(g)=[f,g]$ for all $fg\in A$

$\omega(fX+gY)=f\omega(X)+g\omega(Y)$ for all $f,g\in A$ and all $X,Y\in T^{*}(M)$
For this map to be welldefined, it must respect the relations:
$\omega(f+g)(h)=[f+g,h]=[f,h]+[g,h]=\omega(f)(h)+\omega(g)(h)$ 
$\omega(fg)(h)=[fg,h]=f[g,h]+g[f,h]=f\omega(g)(h)+g\omega(g)(h)$ 
These two equations show that $\omega$ is a welldefined map from the presheaf hence, by general nonsense, a well defined map from the sheaf. The fact that $\omega(f\,dg)$ is a derivation readily follows from the fact that $[,]$ is a derivation in each slot.
Since $[,]$ is nondegenerate, $\omega$ is invertible. Denote its inverse by $\Omega$. Since our manifold is finitedimensional, we may naturally regard $\Omega$ as an element of $T^{*}(M)\otimes T^{*}(M)$. The fact that $\Omega$ is an antisymmetric tensor field (in other words, a 2form) follows from the fact that $\Omega(df)(g)=[f,g]=[g,f]=\Omega(dg)(f)$.
Finally, we will use the Jacobi identity to show that $\Omega$ is closed. If $u,v,w\in T(M)$ then, by a general identity of differential geometry,
$\langle d\Omega,u\wedge v\wedge w\rangle=\langle u,d\langle\Omega,v\wedge w% \rangle\rangle+\langle v,d\langle\Omega,w\wedge u\rangle\rangle+\langle w,d% \langle\Omega,u\wedge v\rangle\rangle$ 
Since this identity is trilinear in $u,v,w$, we can restrict attention to a generating set. Because of the nondegeneracy assumption, vector fields of the form $ad_{f}$ where $f$ is a function form such a set.
By the definition of $\Omega$, we have $\langle\Omega,ad_{f}\wedge ad_{g}\rangle=[f,g]$. Then $\langle ad_{f},d\,\langle\Omega,ad_{g}\wedge ad_{h}\rangle=[f,[g,h]]$ so the Jacobi identity is satisfied.
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